In △ ABC, ∠ a, ∠ B are all acute angles, and Sina = 1 2，tanB= 3, ab = 10, find the area of △ ABC

∵ in ᙽ ABC, ᙽ A and ᙽ B are acute angles, Sina = 1
2，tanB=
3，
∴∠A=30°，∠B=60°，∠C=90°，
∵sinA=a
C=1
2tanB=b
A=
3AB=10，
∴a=1
2c=5，b=
3a=5
3，
∴S△ABC=1
2ab=1
2×5×5
3=25
Three
2．

In △ ABC, ∠ a, ∠ B are all acute angles, and Sina = 1 2，tanB= 3, ab = 10, find the area of △ ABC

In the triangle ABC, the angles a and B are all acute angles, and Sina = 1 / 2, tanb = root 3, D is the midpoint of AB side, and CD = 5. Find the area detail point of triangle ABC

Sina = 1 / 2, a acute angle, so a = 30 degrees, tanb = radical 3, B acute angle, B = 60 degrees
So C = 90 degrees
The oblique side length of a right triangle = the center line on the hypotenuse * 2 = 5 * 2 = 10
So BC = 5
AC = under root sign (100-25) = 5 under root sign 3
Area s = 1 / 2 * 5 * 5 root number 3 = 25 / 2 root number 3

In the acute triangle ABC, ABC is the opposite side of the angle ABC respectively, and the root sign is 3 times that of a is equal to 2 times of C times Sina. If C is equal to the root 7, the area of the triangle ABC is 3 / 2 and the root sign 3 is 3 / 2. Find the value of a + B

In the acute triangle ABC, ABC is the opposite side of the angle ABC respectively, and the root sign is 3 times that of a is equal to 2 times of C times Sina. If C is equal to the root 7, the area of the triangle ABC is 3 / 2 and the root sign 3 is 3 / 2. Find the value of a + B
Analysis: ∵ 3A = 2csina, C = √ 7
∴csinA=√3/2a
∵S(⊿ABC)=3√3/2=1/2bcsinA=1/2ab√3/2==>ab=6
From the sine theorem a / Sina = C / sinc = = > sinc = A / (csina) = √ 3 / 2 = = > C = 60 °
By cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2abcosc
∴a^2+b^2-ab=7=(a+b)^2-3ab=(a+b)^2-18
∴a+b=5

In the triangle ABC, the sides of angles a, B and C are a, B, C respectively, and C = 3 / 4 π, Sina = radical 5 / 5 (1) Find SINB, (2) if C-A = 5-radical 10, find the area of triangle ABC

sinB=sin(A+C)=1/√10
c/a=sinC/sinA=√5/√2
c-a=5-√10
a=2+√10
C=7
S△ABC=(1/2)ac sinB=7/2 + 7√10/10

In the triangle ABC, the opposite sides of the angle A.B.C are a.b.c. it is known that B = the third, Sina = 3 / 5, B = radical 3 Find the value of sinc and the area s of triangle ABC

Sine theorem a / Sina = B / SINB. A = 6 / 5. Sina = 3 / 5, cosa = 4 / 5 (a

In the triangle ABC, given the angle c = 90 degrees, a = 5 times the root sign 3, B = 5, then Sina=

In the triangle ABC, given the angle c = 90 degrees, a = 5 √ 3, B = 5, then Sina=
C = √ (75 + 25) = 10, so Sina = A / C = 5 (√ 3) / 10 = (√ 3) / 2

In △ ABC, if B ^ 2 + C ^ 2-A ^ 2 = root 2BC and SINB / COSC > root 2, then the range of angle c is

Then cosa = (b ^ 2 + C ^ 2-A ^ 2) / (2BC) = √ 2 / 2, a = π / 4.b + C = 3 π / 4, B = 3 π / 4-c.sinb / COSC = sin (3 π / 4-C) / COSC = [(√ 2 / 2) COSC + (√ 2 / 2) sinc] / COSC = √ 2 / 2 + (√ 2 / 2) Tanc > √ 2. Then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tanc > 1, then Tan C > 1, then Tanc > 1, then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tan C > 1, then Tan C so the value range of C is (π / 4, π / 2)

In the triangle ABC, if SINB / COSC > root 2, find the value range of C

Firstly, SINB / COSC > 0, because SINB > 0, COSC > 0, C < 90 °;
sinB/cosC>√2,
sinB＞√2cosC,
Due to SINB ≤ 1,
So COSC < 2 / 2,
That is, C > 45 °,
While C < 90 °,
Therefore, the value range of C is (45 ° and 90 ° respectively)

In triangle ABC, let the opposite sides of triangle a, B and C be a, B and C respectively, and cosC:cosB=3a-c : B 1, find the value of SINB 2, if B = 4 times the root sign 2, In triangle ABC, let the opposite sides of triangle a, B and C be a, B and C respectively, and cosC:cosB=3a-c : B 1, find the value of SINB 2, if B = 4 times the root sign 2, and a = C, find the area of triangle ABC

cosC:cosB=3a-c : B using the sine theorem cosC:cosB= (3sinA-sinC):sinBcosCsinB=3sinAcosB-cosBsinCsin(B+C)=3sinAcosBsinA=3sinAcosBcosB=1/3sinB=2√2/3b²=a²+c²-2accosB32=2a²-2a²/3a²=24S=(1...

In △ ABC, ∠ a, ∠ B are all acute angles, and Sina = 1 2，tanB= 3, ab = 10, find the area of △ ABC